Is $cot^{-1}(-cot x)$ equal to $-x$?
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2
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I was just wondering: We know that $cot ^{-1}(cot x) = x$. But what is the algebraic value of $cot^{-1}(-cot x)$? Is it $-x$?
I am just getting a little muddled up with my trigonometric identities. I would appreciate some clarification.
trigonometry
edited Sep 24 at 16:17
Blue
47.3k870149
asked Sep 24 at 16:14
Aurora Borealis
850414
add a comment |
up vote
2
down vote
favorite
I was just wondering: We know that $cot ^{-1}(cot x) = x$. But what is the algebraic value of $cot^{-1}(-cot x)$? Is it $-x$?
I am just getting a little muddled up with my trigonometric identities. I would appreciate some clarification.
trigonometry
edited Sep 24 at 16:17
Blue
47.3k870149
asked Sep 24 at 16:14
Aurora Borealis
850414
It cannot be for any $x$: $text{arccot}(x)$ is a bounded function while $x$ is not.
– Jack D'Aurizio
Sep 24 at 18:26
Similarly $arctantan x$ is a sawtooth wave, not the identity function. But course they are the same thing over $(-pi/2,pi/2)$.
– Jack D'Aurizio
Sep 24 at 18:27
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was just wondering: We know that $cot ^{-1}(cot x) = x$. But what is the algebraic value of $cot^{-1}(-cot x)$? Is it $-x$?
I am just getting a little muddled up with my trigonometric identities. I would appreciate some clarification.
trigonometry
edited Sep 24 at 16:17
Blue
47.3k870149
asked Sep 24 at 16:14
Aurora Borealis
850414
I was just wondering: We know that $cot ^{-1}(cot x) = x$. But what is the algebraic value of $cot^{-1}(-cot x)$? Is it $-x$?
I am just getting a little muddled up with my trigonometric identities. I would appreciate some clarification.
trigonometry
trigonometry
edited Sep 24 at 16:17
Blue
47.3k870149
asked Sep 24 at 16:14
Aurora Borealis
850414
edited Sep 24 at 16:17
Blue
47.3k870149
asked Sep 24 at 16:14
Aurora Borealis
850414
edited Sep 24 at 16:17
Blue
47.3k870149
edited Sep 24 at 16:17
Blue
47.3k870149
edited Sep 24 at 16:17
Blue
47.3k870149
47.3k870149
asked Sep 24 at 16:14
Aurora Borealis
850414
asked Sep 24 at 16:14
Aurora Borealis
850414
asked Sep 24 at 16:14
Aurora Borealis
850414
850414
It cannot be for any $x$: $text{arccot}(x)$ is a bounded function while $x$ is not.
– Jack D'Aurizio
Sep 24 at 18:26
Similarly $arctantan x$ is a sawtooth wave, not the identity function. But course they are the same thing over $(-pi/2,pi/2)$.
– Jack D'Aurizio
Sep 24 at 18:27
add a comment |
It cannot be for any $x$: $text{arccot}(x)$ is a bounded function while $x$ is not.
– Jack D'Aurizio
Sep 24 at 18:26
Similarly $arctantan x$ is a sawtooth wave, not the identity function. But course they are the same thing over $(-pi/2,pi/2)$.
– Jack D'Aurizio
Sep 24 at 18:27
It cannot be for any $x$: $text{arccot}(x)$ is a bounded function while $x$ is not.
– Jack D'Aurizio
Sep 24 at 18:26
It cannot be for any $x$: $text{arccot}(x)$ is a bounded function while $x$ is not.
– Jack D'Aurizio
Sep 24 at 18:26
Similarly $arctantan x$ is a sawtooth wave, not the identity function. But course they are the same thing over $(-pi/2,pi/2)$.
– Jack D'Aurizio
Sep 24 at 18:27
Similarly $arctantan x$ is a sawtooth wave, not the identity function. But course they are the same thing over $(-pi/2,pi/2)$.
– Jack D'Aurizio
Sep 24 at 18:27
add a comment |
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
HINT
Remember that the cotangent function is odd:
begin{align*}
cot(x) = frac{cos(x)}{sin(x)} Longrightarrow cot(-x) = frac{cos(-x)}{sin(-x)} = -frac{cos(x)}{sin(x)} = -cot(x)
end{align*}
answered Sep 24 at 16:18
APC89
1,747318
Ok so it essntially means then $cot ^{-1} (-cot x) = x ? $ I get it now Thank you.
– Aurora Borealis
Sep 24 at 16:21
1
No, you have $cot^{-1}(-cot(x)) = cot^{-1}(cot(-x)) = -x$.
– APC89
Sep 24 at 16:23
add a comment |
up vote
2
down vote
$$ cot^{-1}(cot(x)) = x$$
is not actually a trigonometric 'identity', it is a result of inverse function, $ f(f^{-1}(y)) = y $
$$ cot( cot^{-1}(cot(x)) ) = cot(x) implies cot^{-1}(cot(x)) = x $$
by @APC89's answer:
$$ cot( cot^{-1}(-cot(x)) ) = -cot(x) = cot(-x) $$
so we have
$$ cot^{-1}(-cot(x))= -x $$
answered Sep 24 at 16:32
Arief
1,3601622
add a comment |
up vote
2
down vote
Yes, it is. Working backwards a bit, we can confirm this.
Remember that $cos (-x)$ = $cos (x)$ and $sin (-x)$ = $-sin (x)$.
$$cot^{-1}(-cot x) = cot^{-1}bigg(frac{cos (-x)}{sin (-x)}bigg) = cot^{-1} (cot (-x)) = -x$$
answered Sep 24 at 16:41
KM101
3,950417
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
HINT
Remember that the cotangent function is odd:
begin{align*}
cot(x) = frac{cos(x)}{sin(x)} Longrightarrow cot(-x) = frac{cos(-x)}{sin(-x)} = -frac{cos(x)}{sin(x)} = -cot(x)
end{align*}
answered Sep 24 at 16:18
APC89
1,747318
Ok so it essntially means then $cot ^{-1} (-cot x) = x ? $ I get it now Thank you.
– Aurora Borealis
Sep 24 at 16:21
1
No, you have $cot^{-1}(-cot(x)) = cot^{-1}(cot(-x)) = -x$.
– APC89
Sep 24 at 16:23
add a comment |
up vote
6
down vote
accepted
HINT
Remember that the cotangent function is odd:
begin{align*}
cot(x) = frac{cos(x)}{sin(x)} Longrightarrow cot(-x) = frac{cos(-x)}{sin(-x)} = -frac{cos(x)}{sin(x)} = -cot(x)
end{align*}
answered Sep 24 at 16:18
APC89
1,747318
Ok so it essntially means then $cot ^{-1} (-cot x) = x ? $ I get it now Thank you.
– Aurora Borealis
Sep 24 at 16:21
1
No, you have $cot^{-1}(-cot(x)) = cot^{-1}(cot(-x)) = -x$.
– APC89
Sep 24 at 16:23
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
HINT
Remember that the cotangent function is odd:
begin{align*}
cot(x) = frac{cos(x)}{sin(x)} Longrightarrow cot(-x) = frac{cos(-x)}{sin(-x)} = -frac{cos(x)}{sin(x)} = -cot(x)
end{align*}
answered Sep 24 at 16:18
APC89
1,747318
HINT
Remember that the cotangent function is odd:
begin{align*}
cot(x) = frac{cos(x)}{sin(x)} Longrightarrow cot(-x) = frac{cos(-x)}{sin(-x)} = -frac{cos(x)}{sin(x)} = -cot(x)
end{align*}
answered Sep 24 at 16:18
APC89
1,747318
answered Sep 24 at 16:18
APC89
1,747318
answered Sep 24 at 16:18
APC89
1,747318
answered Sep 24 at 16:18
APC89
1,747318
1,747318
Ok so it essntially means then $cot ^{-1} (-cot x) = x ? $ I get it now Thank you.
– Aurora Borealis
Sep 24 at 16:21
1
No, you have $cot^{-1}(-cot(x)) = cot^{-1}(cot(-x)) = -x$.
– APC89
Sep 24 at 16:23
add a comment |
Ok so it essntially means then $cot ^{-1} (-cot x) = x ? $ I get it now Thank you.
– Aurora Borealis
Sep 24 at 16:21
1
No, you have $cot^{-1}(-cot(x)) = cot^{-1}(cot(-x)) = -x$.
– APC89
Sep 24 at 16:23
Ok so it essntially means then $cot ^{-1} (-cot x) = x ? $ I get it now Thank you.
– Aurora Borealis
Sep 24 at 16:21
Ok so it essntially means then $cot ^{-1} (-cot x) = x ? $ I get it now Thank you.
– Aurora Borealis
Sep 24 at 16:21
1
1
No, you have $cot^{-1}(-cot(x)) = cot^{-1}(cot(-x)) = -x$.
– APC89
Sep 24 at 16:23
No, you have $cot^{-1}(-cot(x)) = cot^{-1}(cot(-x)) = -x$.
– APC89
Sep 24 at 16:23
add a comment |
up vote
2
down vote
$$ cot^{-1}(cot(x)) = x$$
is not actually a trigonometric 'identity', it is a result of inverse function, $ f(f^{-1}(y)) = y $
$$ cot( cot^{-1}(cot(x)) ) = cot(x) implies cot^{-1}(cot(x)) = x $$
by @APC89's answer:
$$ cot( cot^{-1}(-cot(x)) ) = -cot(x) = cot(-x) $$
so we have
$$ cot^{-1}(-cot(x))= -x $$
answered Sep 24 at 16:32
Arief
1,3601622
add a comment |
up vote
2
down vote
$$ cot^{-1}(cot(x)) = x$$
is not actually a trigonometric 'identity', it is a result of inverse function, $ f(f^{-1}(y)) = y $
$$ cot( cot^{-1}(cot(x)) ) = cot(x) implies cot^{-1}(cot(x)) = x $$
by @APC89's answer:
$$ cot( cot^{-1}(-cot(x)) ) = -cot(x) = cot(-x) $$
so we have
$$ cot^{-1}(-cot(x))= -x $$
answered Sep 24 at 16:32
Arief
1,3601622
add a comment |
up vote
2
down vote
up vote
2
down vote
$$ cot^{-1}(cot(x)) = x$$
is not actually a trigonometric 'identity', it is a result of inverse function, $ f(f^{-1}(y)) = y $
$$ cot( cot^{-1}(cot(x)) ) = cot(x) implies cot^{-1}(cot(x)) = x $$
by @APC89's answer:
$$ cot( cot^{-1}(-cot(x)) ) = -cot(x) = cot(-x) $$
so we have
$$ cot^{-1}(-cot(x))= -x $$
answered Sep 24 at 16:32
Arief
1,3601622
$$ cot^{-1}(cot(x)) = x$$
is not actually a trigonometric 'identity', it is a result of inverse function, $ f(f^{-1}(y)) = y $
$$ cot( cot^{-1}(cot(x)) ) = cot(x) implies cot^{-1}(cot(x)) = x $$
by @APC89's answer:
$$ cot( cot^{-1}(-cot(x)) ) = -cot(x) = cot(-x) $$
so we have
$$ cot^{-1}(-cot(x))= -x $$
answered Sep 24 at 16:32
Arief
1,3601622
answered Sep 24 at 16:32
Arief
1,3601622
answered Sep 24 at 16:32
Arief
1,3601622
answered Sep 24 at 16:32
Arief
1,3601622
1,3601622
add a comment |
add a comment |
up vote
2
down vote
Yes, it is. Working backwards a bit, we can confirm this.
Remember that $cos (-x)$ = $cos (x)$ and $sin (-x)$ = $-sin (x)$.
$$cot^{-1}(-cot x) = cot^{-1}bigg(frac{cos (-x)}{sin (-x)}bigg) = cot^{-1} (cot (-x)) = -x$$
answered Sep 24 at 16:41
KM101
3,950417
add a comment |
up vote
2
down vote
Yes, it is. Working backwards a bit, we can confirm this.
Remember that $cos (-x)$ = $cos (x)$ and $sin (-x)$ = $-sin (x)$.
$$cot^{-1}(-cot x) = cot^{-1}bigg(frac{cos (-x)}{sin (-x)}bigg) = cot^{-1} (cot (-x)) = -x$$
answered Sep 24 at 16:41
KM101
3,950417
add a comment |
up vote
2
down vote
up vote
2
down vote
Yes, it is. Working backwards a bit, we can confirm this.
Remember that $cos (-x)$ = $cos (x)$ and $sin (-x)$ = $-sin (x)$.
$$cot^{-1}(-cot x) = cot^{-1}bigg(frac{cos (-x)}{sin (-x)}bigg) = cot^{-1} (cot (-x)) = -x$$
answered Sep 24 at 16:41
KM101
3,950417
Yes, it is. Working backwards a bit, we can confirm this.
Remember that $cos (-x)$ = $cos (x)$ and $sin (-x)$ = $-sin (x)$.
$$cot^{-1}(-cot x) = cot^{-1}bigg(frac{cos (-x)}{sin (-x)}bigg) = cot^{-1} (cot (-x)) = -x$$
answered Sep 24 at 16:41
KM101
3,950417
edited Nov 27 at 17:14
answered Sep 24 at 16:41
KM101
3,950417
answered Sep 24 at 16:41
KM101
3,950417
answered Sep 24 at 16:41
KM101
3,950417
3,950417
add a comment |
add a comment |
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It cannot be for any $x$: $text{arccot}(x)$ is a bounded function while $x$ is not.
– Jack D'Aurizio
Sep 24 at 18:26
Similarly $arctantan x$ is a sawtooth wave, not the identity function. But course they are the same thing over $(-pi/2,pi/2)$.
– Jack D'Aurizio
Sep 24 at 18:27